Discover the Pi Paradox: The Infinite Perimeter of a Circle with a Diameter of 1

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Discussion Overview

The discussion centers around a mathematical exploration of the relationship between the perimeter of shapes inscribed within a circle of diameter 1 and the value of pi. Participants examine the implications of modifying a square into a cross shape and further into a curve, questioning whether this leads to a conclusion that pi equals 4 or 2.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant proposes that by continuously indenting the corners of a square inscribed in a circle, the perimeter remains 4, suggesting that this leads to the conclusion that pi equals 4.
  • Another participant challenges this claim, stating that the perimeter of the square would approximate the circumference of the circle, indicating that further mathematical work is needed to establish the value of pi.
  • A different participant argues that the original claim does not prove the length of the circle is 4, but rather illustrates a concept about limits, suggesting that the limiting process does not yield the expected result.
  • Another participant contradicts the initial claim by asserting that pi equals 2, using an argument based on approximating the area under the curve of the circle with rectangles, leading to a total perimeter of 2 across all quadrants.
  • One participant notes that this problem has been extensively discussed in another math forum, implying ongoing debate and interest in the topic.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the relationship between the perimeter of the shapes and the value of pi, with no consensus reached on the validity of the claims presented.

Contextual Notes

Participants reference different mathematical approaches and concepts, such as limits and integral approximations, without resolving the underlying assumptions or mathematical steps involved in their arguments.

jamester234
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Suppose you have a circle with a diameter of 1. If you draw a square with all four sides touching the circle, the perimeter of the square is 4. Now suppose you indent each corner of the square so that they all touch the circle-this will make a cross shape, and the perimeter of it is still 4. Now suppose you indent each corner of the cross so that they again all touch the circle. You can see now that the square is becoming more like a circle, and yet the pereimeter is still 4. This can be done to infinity, therefore, pi equals 4!
 
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jamester234 said:
Suppose you have a circle with a diameter of 1. If you draw a square with all four sides touching the circle, the perimeter of the square is 4. Now suppose you indent each corner of the square so that they all touch the circle-this will make a cross shape, and the perimeter of it is still 4. Now suppose you indent each corner of the cross so that they again all touch the circle. You can see now that the square is becoming more like a circle, and yet the pereimeter is still 4. This can be done to infinity, therefore, pi equals 4!

the perimeter of your square would become roughly equal to the circumference if your circle. you still got some math to do before you get pi.
 
Nope. You have not proved the length of the circle (your limiting curve) is 4. You have proved, instead, that the length of the limit is not equal to the limit of the lengths. No need to use a circle to prove this: For example, you can use stairsteps converging to a sloping line segment.
 
That demonstration is obviously wrong because it is contradicted by the proof that \pi = 2. Consider a circle of diameter 1 with its center at the origin and look at the part of it that lies in the 1st quadrant. Divide up the area under the curve of the circle into small rectangles in the manner that people do when they approximate integrals. The tops of these rectangles become closer and closer to the curve as their widths approach zero. The total of the lengths of the top of the rectangles is always 1/2 no matter how small they are. So the total perimeter of the circle is 2 when we sum over all 4 quadrants.
 
This exact problem was just thrashed to death in another math forum on here.
 

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